This is basically the square version of the top in reply #81 .

**Remaining question Nr.1:** Can this way of making offset tops be generalized as the case with the wire frame that ta0 found to be possible for almost anything?

**Remaining question Nr.2:** What would be an elegant prove for the fact that the position of the center of mass is where I put that red dot? I just did a straightforward calculation with the "lever rule" and was happy that the outcome was again the Golden Ratio, but a prove for that, that would be equally elegant as I think the resulting top is, would be different......

[1] Maybe this can count. If you have a figure enclosed in a square, with it's centroid on the center of the square, and place stretched copies on two sides, the center of mass will fall on the corner of the square if the stretch factor is the Golden Ratio:

I don't even think the figure has to have circular symmetry: it could look different when stretched along the x and y axis. This is because what matters is that its center remains in the center of the rectangle and its area increases proportional to the stretch factor.

One problem for making a model is that the resulting center of mass can be outside the figure. You can fix this by replacing the original figure with one with the same area and also centered on the square, like I did on the right.

[2] I think the most direct derivation for the squares is to consider that the lever of the center of mass of the L shape and the one of the missing square should equate at the center of the big square. Mass small square = 1, Mass missing square = x

^{2}, lever arm small square = (x-1), lever arm missing square = 1 (where I took the lever in the x direction to simplify). Equating everything you get the Golden Ratio equation.